2020 Jan Gold Problem 1 Time is Mooney

Official Problem Statement

Time is Mooney

Problem Statement

The goal of this problem is to determine the minimum amount of time needed to complete a sequence of tasks.

You are given a list of N tasks, each of which takes a certain amount of time to complete. You can only work on one task at a time, and you must complete all tasks in the given order. Additionally, you have a limited amount of time, T, to complete all the tasks.

Your task is to determine the minimum amount of time needed to complete all the tasks.

Solution

The solution to this problem can be solved using dynamic programming.

We create an array dp[N+1] where dp[i] represents the minimum amount of time needed to complete the first i tasks.

We can then fill the array using the following recurrence relation:

dp[i] = min(dp[i], dp[i-1] + time[i])

Where time[i] is the time required to complete the i-th task.

We can then iterate through the array and find the minimum amount of time needed to complete all tasks.

The following is a C++ implementation of the solution:

int N; // number of tasks
int T; // total time available
int time[N+1]; // time required to complete each task
int dp[N+1]; // dp[i] represents the minimum amount of time needed to complete the first i tasks

dp[0] = 0;

for (int i = 1; i <= N; i++) {
    dp[i] = T + 1; // set dp[i] to a value greater than T
    for (int j = 0; j < i; j++) {
        dp[i] = min(dp[i], dp[j] + time[i]);
    }
}

int ans = T + 1;
for (int i = 0; i <= N; i++) {
    ans = min(ans, dp[i]);
}

if (ans > T) {
    cout << "Not possible" << endl;
} else {
    cout << ans << endl;
}